Question

If `(log)_k xdot(log)_5k=(log)_x5,k!=1,k >0,` then `x` is equal to `k` (b) 1/5 (c) 5 (d) none of these
A. k
B. `1//5`
C. 5
D. none of these

Answer 1

Correct Answer - B::C
`log_(k) x * log_(5) k=log_(x) 5`
` or(log x log k)/(log k log 5) = log_(x) 5 `
` or (log x )/(log 5) = log _(x) 5`
` or log_(5) x = 1/(log_(5) x)`
` or (log_(5)x)^(2) = 1`
` or log_(5) x = pm 1`
` rArr x = 5^(pm 1) `
` rArr x = 1/5 , 5`